Is it true that if $\mbox{Ext}^{1}_{A}(P,A/I)=0 \ \forall I$ then P is projective? Similar statements are true for flat and injective modules, but I'm beginning to suspect that projective modules cannot be characterized soley by ideals.

I can see the first implication, but not the second. Does Ext commute with direct limit?
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ashpoolMay 27 '10 at 2:14

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If $P$ is finitely generated, some finite rank free module $F$ surjects onto $P$. Since $A$ is Noetherian, the kernel is also finitely generated. From your condition, the induced map from $Hom(P,F)$ to $Hom(P,P)$ is surjective. This is equivalent to $P$ being projective; see Prop. A3.1 in Eisenbud.
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Steve DMay 27 '10 at 9:20

Thanks! I have never seen that property before.
–
ashpoolMay 27 '10 at 20:45

By the way, the proof also seems to work if I drop the Noetherian condition and impose P to be finitely presented.
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ashpoolMay 27 '10 at 21:39

When $A=\mathbb Z$ the condition is equivalent to $\mathrm{Ext}^1_{\mathbb Z}(A,\mathbb Z)=0$ and the problem as to whether this implies that $A$ is free is the Whitehead problem and was shown by Shelah to be undecidable in ZFC (standard set theory). Hence there is at least one ring for which the problem is difficult.

Thanks for the comment. I don't quite see the equivalence of the two statements? Do you think you can give me some tips?
–
ashpoolMay 27 '10 at 22:14

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As we always have $\mathrm{Ext}^2(A,M)=0$, the short exact sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/n\to0$ gives that $\mathrm{Ext}^1(A,\mathbb Z/n)=0$ for all $n$ if $\mathrm{Ext}^1(A,\mathbb Z)=0$.
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Torsten EkedahlMay 31 '10 at 18:22

I claim that $M = R/\mathfrak{m}$ is a counterexample, i.e. $\operatorname{Ext}_R^1(M,R/I) = 0$ for all ideals $I$ of $R$, yet $M$ is not a projective $R$-module.

Proof:

$M$ is not projective: If $M$ were projective, then $0 \to \mathfrak{m} \to R \to M \to 0$ would split, hence $\mathfrak{m}$ would be a quotient of $R$, so in particular it could be generated by one element. However $\mathfrak{m}$ is not a finitely generated ideal.

Here $\operatorname{Ext}^1(R,R/I) = 0$ since $R$ is projective. So it is enough to prove that $R/I = \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I)$ is surjective.

The ideals of $R$ are easy to describe: If $c \in \mathbb{R}_{\ge 0}$, then let $I_{\ge c} = (X^c)$ and $I_{>c} = (X^r ; \, r > c)$. Then every nonzero ideal is of the form $I_{\ge c}$ or $I_{>c}$. In particular $\mathfrak{m} = I_{>0}$.

If $I=0$:
we need that $R = \operatorname{Hom}(R,R) \to \operatorname{Hom}(\mathfrak{m}, R)$ is surjective (in fact it is bijective).
This is not hard to check (for $\varphi \in \operatorname{Hom}(\mathfrak{m}, R)$,
show that $\varphi(X^r) \in I_{\ge r}$, and $X^{-r} \varphi(X^r) \in R$ is independent of $r>0$).

If $I=I_{\ge c}$:
Let $\varphi \colon \mathfrak{m} \to R/I_{\ge c}$ be a homomorphism.
We need to show that there is an $h \in R$ such that $\varphi(f) = f h + I_{\ge c}$.
To prove this, look at $\varphi(X^r)$ and take $r \to 0$.
If $r<c$, then $\varphi(X^r)$ will determine $h$ modulo $I_{\ge c-r}$.
To finish the proof, we need that if $E \subseteq [0, c)$ such that $E \cap [0,d)$ is well-ordered for all $d<c$, then $E$ is also well-ordered.

Well known but not provable as illustrated in another answer. Maybe it should go as an answer to the common false beliefs in mathematics question that has just resurfaced again.
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Simon WadsleyMar 3 '11 at 18:46